How Tartaglia Solved the Cubic Equation - Tartaglia's Poem

Tartaglia divulged to Hieronimo Cardano (1501–1576) the solution of the three cubic equations without the quadratic term on March 25, 1539, in Cardano's house in Milano in the form of a famous poem (translated here by the author):

01) When the cube with the cose beside it <\(x^3+px\)>
02) Equates itself to some other whole number, <\(=q\)>
03) Find two other, of which it is the difference. <\(u-v=q\)>
04) Hereafter you will consider this customarily
05) That their product always will be equal <\(uv=\)>
06) To the third of the cube of the cose net. <\(\frac{p^3}{3}\), instead of \((\frac{p}{3})^3\)>
07) Its general remainder then
08) Of their cube sides , well subtracted, <\({\sqrt[3]u}-{\sqrt[3]v}\)>
09) Will be the value of your principal unknown. <\(=x\)>
10) In the second of these acts,
11) When the cube remains solo , <\(x^3=px+q\)>
12) You will observe these other arrangements:
13) Of the number <q> you will quickly make two such parts, <\(q=u+v\)>
14) That the one times the other will produce straightforward <\(uv=\)>
15) The third of the cube of the cose in a multitude, <\(\frac{p^3}{3}\), instead of \((\frac{p}{3})^3\)>
16) Of which then, per common precept,
17) You will take the cube sides joined together. <\({\sqrt[3]u}+{\sqrt[3]v}\) 
18) And this sum will be your concept. <\(=x\)>
19) The third then of these our calculations <\(x^3+q=px\)>
20) Solves itself with the second, if you look well after,
21) That by nature they are quasi conjoined.
22) I found these, & not with slow steps,
23) In thousand five hundred, four and thirty
24) With very firm and strong foundations
25) In the city girded around by the sea.

Of course, in the English translation the lines do not rhyme as they do in the Italian original. There the poem is made highly artistic. Tartaglia chose a kind of poem—a so-called capitolo (the same word also means chapter, and type of equation)—constructed in a very complicated manner. A capitolo poem is written in the so-called terza rima (Italian: third rhyme), the structure of which consists of groups of three lines of verse (called tercet in English), lines 01 to 03, 04 to 06, 07 to 09, etc., where the first and the third line rhyme. The peculiarity of this kind of poems is that the middle line of a tercet rhymes with the first and the third line of the next triplet. If the lines are represented by letters, and the same letter means that these lines rhyme, the pattern of a terza rima is aba, bcb, cdc, ded, and so on. (The poem in the original Italian appears below.)

The necessity to follow this scheme led to two misleading instructions in line 6 and 15. In line 6 the wording says "the third of the cube of the cose net." The "cose net" in this case is the coefficient p. "The third of the cube," therefore, means \(\frac{p^3}{3}\). That is wrong. Correctly it should be "the cube of the third of the unknown," \((\frac{p}{3})^3\), and that is why this passage confused Cardano. In a letter from 9 April he implored Tartaglia for help to solve the equation .1.cubo piu.3.cose equal à 10. Tartaglia answered on April 23, 1539, and gave the explanatory instructions we describe in the previous section.

Lines 19, 20, and 21 mean that you solve this type of cubic equation, our C), with the second type B) \(x^3=px+q\). The second type \(x^3=px+q\) and the third type \(x^3+q=px\) with the same p and q, as was said before, have numerically the same three roots but with opposite signs. Type B) always has a positive root x₁. Consequently, the root of the corresponding equation of the third type was negative, and, therefore, rejected.

Line 23: In the Republic of Venice until its end in 1797 the year did not begin on January 1 as today but officially on March 1. Therefore, February 12 and 13 of our year 1535 were still in the Venetian year 1534.